2019年3月19日
摘要: I hate geometric problems…... java class Solution { public int minAreaRect(int[][] points) { Map xmap = new HashMap(); Map ymap = new HashMap(); Set s 阅读全文
posted @ 2019-03-19 21:24 王 帅 阅读(93) 评论(0) 推荐(0) 编辑
摘要: dont know why, but I prefer to use for instead of while. 阅读全文
posted @ 2019-03-19 21:23 王 帅 阅读(106) 评论(0) 推荐(0) 编辑
摘要: Keep two int variable to record min and max value, when encounter I, use max++ value, on the contrary, use min . At last, add (min + 1) to all element 阅读全文
posted @ 2019-03-19 21:22 王 帅 阅读(57) 评论(0) 推荐(0) 编辑
摘要: It is marked as easy, and it's really very easy. 阅读全文
posted @ 2019-03-19 21:20 王 帅 阅读(71) 评论(0) 推荐(0) 编辑
摘要: Just use a stack to simulate the push and pop process While stack is empty or peek does not equal to popped current value, push value into stack. Whil 阅读全文
posted @ 2019-03-19 21:19 王 帅 阅读(140) 评论(0) 推荐(0) 编辑